Question: Solve for $x$ : $2\sqrt{x} + 3 = 8\sqrt{x} + 4$
Explanation: Subtract $2\sqrt{x}$ from both sides: $(2\sqrt{x} + 3) - 2\sqrt{x} = (8\sqrt{x} + 4) - 2\sqrt{x}$ $3 = 6\sqrt{x} + 4$ Subtract $4$ from both sides: $3 - 4 = (6\sqrt{x} + 4) - 4$ $-1 = 6\sqrt{x}$ Divide both sides by $6$ $\frac{-1}{6} = \frac{6\sqrt{x}}{6}$ Simplify. $-\dfrac{1}{6} = \sqrt{x}$ The principal root of a number cannot be negative. So, there is no solution.